Multivariable limit with polar coordinates Polar coordinates do not reveal the behaviour of $f(x,y)$ when studying
$$ \lim_{x^2 + y^2 \to \infty} \frac {xy}{e^{x^2y^2}} $$
In polar coordinates we have
$$ \lim_{r^2 \to \infty} \frac 12 \frac { r^2 \sin (2 \varphi ) }{ e^{ \frac 14 r^4 \sin^2 (2\varphi) } } $$
which goes to zero independently of $\varphi$. However by letting $y = \frac 1x$, we clearly see that the limit cannot exist. Why didn't polar coordinates work out and what other way than  letting $y = \frac 1x$ can we show that it does not exist?
 A: But the issue is why polar coordinates can lead to error. In fact this is a recurrent misunderstanding. To explain it I turn to limits when $(x,y)\to(0,0)$ (the same thing up to an inversion). The condition
for all $\theta$ there exist $\lim_{\rho\to0}f(\rho\cos\theta,\rho\sin\theta)$ and does not depend on $\theta$,
only says that all limits along lines through the origin exist and coincide. This is not enough for the limit to exist, there could be different limits along paraboles, for instance (hyperbolas in a comment above). What is equivalent to the existence of a limit, say $L$, is
for all $\varepsilon>0$ there is $\delta>0$ such that if $\ 0<\rho<\delta$,
then for all $0\le\theta<2\pi$ we have $\ |f(\rho\cos\theta,\rho\sin\theta)-L|<\varepsilon.$
In other words, what must be independent of $\theta$ is the rate of convergency $\delta$, besides the limit $l$. 
A: Consider the two hyperbolas
$$\gamma_\pm:\qquad y=\pm{1\over x}\quad(x>0)\ .$$
When $(x,y)\in\gamma_+$ then $f(x,y)={1\over e}$, and when $(x,y)\in\gamma_-$ then $f(x,y)=-{1\over e}$. Since both hyperbolas extends to infinity there are are points $(x,y)$ with $x^2+y^2$ arbitrarily large and $f(x,y)={1\over e}$, and there are other such points with $f(x,y)=-{1\over e}$. It follows that the envisaged limit cannot exist.
A: Idea: you can study the auxiliar function $z\mapsto z/\exp z^2$. Obviously, $z=xy$.
EDIT: do the limits on the subsets $x=0$ (or $y=0$, or $x=y$) and
$y={1\over\sqrt2 \,x}.$
